Evaluate the improper integral if it exists. $\int^{\infty}_{1}e^x\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $e$ (Choice C) C $\dfrac{1}{e}$ (Choice D) D The improper integral diverges.
Solution: First, let's rewrite the improper integral: $\int^{\infty}_{1}e^x\,dx=\lim_{b\to\infty}\int_{1}^{b}e^x\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_{2}^{\infty}\dfrac1{\sqrt x}\,dx}&=\lim_{b\to\infty}\int_{1}^{b}e^x\,dx\\ \\ \\ &=\lim_{b\to\infty}\Big[e^x\Big]_{1}^b\\ \\ \\ &=\lim_{b\to\infty}\left(e^b-e^1\right)\\ \\ \\ &=\lim_{b\to\infty}\left(e^b\right)-\lim_{b\to\infty}\left(e^1\right)\\\\\\ &=\infty-e\\ \\ &=\infty \end{aligned}$ The answer: The improper integral diverges.